A First Order Extension of Stlmarcks Method

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A First Order Extension of Stålmarcks Method

• Magnus Björk
• Chalmers University of Technology
• Gothenburg, Sweden

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Overview

• Stålmarcks method (in propositional logic)
• Features of FOL version
• Soundness completeness
• Implementation benchmarks

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Stålmarcks Method

• Tableaux like
• Operates on propositional formulae
• Successful in industrial applications
• Used in Prover Plug In
• Non-branching rules KE KI
• Dilemma rule branch and merge

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Example in Propositional Logic
P?Q Q?P (P?Q)?R
Input formulae
5
Example in Propositional Logic
P?Q Q?P (P?Q)?R
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Example in Propositional Logic
P?Q Q?P (P?Q)?R
P
P
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Example in Propositional Logic
P?Q Q?P (P?Q)?R
P Q P?Q R
P
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Example in Propositional Logic
P?Q Q?P (P?Q)?R
P Q P?Q R
P Q P?Q R
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Example in Propositional Logic
P?Q Q?P (P?Q)?R
P Q P?Q R
P Q P?Q R
P?Q R
The intersection of the two branches
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Structure of a proof
0-hard problem No dilemma rule applications
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Structure of a proof
1-hard problem No nested dilemmas
X
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Structure of a proof
2-hard problem Maximal nesting level of 2
X
X
X
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Our Version of First Order Logic

• Rigid and universal variables
• Universal variables implicitly universally
  quantified can be instantiated with anything
  during proof
• Rigid variables similar to constants may not be
  instantiated

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Example in FOL
"x,y.(P(x,y)?Q(x,y)) "x,y.(Q(x,c)?P(x,c)) "x,y.((P
(x,y)?Q(x,y))?R(x,y))
Input formulae
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Example in FOL
"x,y.(P(x,y)?Q(x,y)) "x,y.(Q(x,c)?P(x,c)) "x,y.((P
(x,y)?Q(x,y))?R(x,y)) P(x,y)?Q(x,y) Q(x,c)?P(x,c)
(P(x,y)?Q(x,y))?R(x,y)
Input formulae
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Example in FOL
Input formulae
... P(x,y)?Q(x,y) Q(x,c)?P(x,c) (P(x,y)?Q(x,y))?R(
x,y)
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Example in FOL
... P(x,y)?Q(x,y) Q(x,c)?P(x,c) (P(x,y)?Q(x,y))?R(
x,y)
Future dilemma formulae P(U,V),
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Example in FOL
... P(x,y)?Q(x,y) Q(x,c)?P(x,c) (P(x,y)?Q(x,y))?R(
x,y)
Future dilemma formulae
ØP(U,V)
P(U,V)
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Example in FOL
... P(x,y)?Q(x,y) Q(x,c)?P(x,c) (P(x,y)?Q(x,y))?R(
x,y)
Future dilemma formulae
ØP(U,V)
P(U,V) Q(U,V) P(U,V)?Q(U,V) R(U,V)
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Example in FOL
... P(x,y)?Q(x,y) Q(x,c)?P(x,c) (P(x,y)?Q(x,y))?R(
x,y)
Future dilemma formulae , P(U,c)
ØP(U,V) Suggestion V?c
P(U,V) Q(U,V) P(U,V)?Q(U,V) R(U,V)
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Example in FOL
... P(x,y)?Q(x,y) Q(x,c)?P(x,c) (P(x,y)?Q(x,y))?R(
x,y)
Future dilemma formulae , P(U,c)
ØP(U,V) Suggestion V?c
P(U,V) Q(U,V) P(U,V)?Q(U,V) R(U,V)
No common consequences
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Example in FOL
... P(x,y)?Q(x,y) Q(x,c)?P(x,c) (P(x,y)?Q(x,y))?R(
x,y)
Future dilemma formulae , P(U,c)
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Example in FOL
... P(x,y)?Q(x,y) Q(x,c)?P(x,c) (P(x,y)?Q(x,y))?R(
x,y)
Future dilemma formulae P(U,c)

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Example in FOL
... P(x,y)?Q(x,y) Q(x,c)?P(x,c) (P(x,y)?Q(x,y))?R(
x,y)
Future dilemma formulae

ØP(U,c)
P(U,c)
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Example in FOL
... P(x,y)?Q(x,y) Q(x,c)?P(x,c) (P(x,y)?Q(x,y))?R(
x,y)
Future dilemma formulae

ØP(U,c)
P(U,c) Q(U,c) P(U,c)?Q(U,c) R(U,c)
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Example in FOL
... P(x,y)?Q(x,y) Q(x,c)?P(x,c) (P(x,y)?Q(x,y))?R(
x,y)
Future dilemma formulae

ØP(U,c) ØQ(U,c) P(U,c)?Q(U,c) R(U,c)
P(U,c) Q(U,c) P(U,c)?Q(U,c) R(U,c)
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Example in FOL
... P(x,y)?Q(x,y) Q(x,c)?P(x,c) (P(x,y)?Q(x,y))?R(
x,y)
Future dilemma formulae

ØP(U,c) ØQ(U,c) P(U,c)?Q(U,c) R(U,c)
P(U,c) Q(U,c) P(U,c)?Q(U,c) R(U,c)
P(x,c)?Q(x,c) R(x,c)
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Finding Dilemma Formulae

• Begin with general formulae (containing rigid
  variables)
• Find instances when unifications fail due to
  rigid variables
• Redo dilemmas with instantiated dilemma formulae

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Why no Destructive Updates?

• Find general consequences, since rigid variables
  become universal after merge
• Dont have to worry about fairness all
  instances will be tried out eventually

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Logical Intersections

• Ordinary set intersection wasteful on sets of
  first order formulae
• E.g. P(x) n P(c) Ø, but P(c) consequence of
  both sets
• Logical intersection unifies formulae from the
  two sets

P(f(y)) ØP(c) Q(U)
P(x) Q(c)
P(f(y)) Suggestion U?c
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Proof procedure

• Begin with all non-branching rules until a
  certain formula complexity (0-saturation)
• Perform all dilemmas with 0-saturation in
  branches (1-saturation)
• Increase dilemma nesting level
  (2, 3, -saturation)

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Completeness

• Say that a set of formulae is n-saturated if all
  consequences that can be derived with a dilemma
  nesting level of at most n are in the set
• Note that the proof system simulates KE, and
  therefore any unsatisfiable formula set has a
  finite dilemma refutation
• Show that proof procedure produces increasingly
  saturated sets

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Soundness of the Dilemma Rule
Assume A0s A1s A
P(U) ØP(U)
A0 A1 AU?x
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Soundness of the Dilemma Rule
Assume A0s A1s A
P(U) ØP(U)
A0 A1 AU?x
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Soundness of the Dilemma Rule
Assume A0s A1s A
P(U) or ØP(U)
A0 A1 AU?x
for all U
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Soundness of the Dilemma Rule
Assume A0s A1s A
P(U) or ØP(U)
A0 or A1 AU?x
for all U for all U
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Soundness of the Dilemma Rule
Assume A0s A1s A
P(U) or ØP(U)
A0 or A1 A0 v A1 AU?x
for all U for all U for all U
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Soundness of the Dilemma Rule
Assume A0s A1s A
P(U) or ØP(U)
A0 or A1 A0 v A1 A v
A AU?x
for all U for all U for all U for all U
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Soundness of the Dilemma Rule
Assume A0s A1s A
P(U) or ØP(U)
A0 or A1 A0 v A1 A v
A A AU?x
for all U for all U for all U for all
U for all U
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Implementation Dilemma

• Made during 2004
• No equality support, lots of room for improvement
• Participated in CASC-J2 (2004)

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Benchmarks

• TPTP 2.7.0
• Formulae w/o equality 192/305 63
• Formulae with equality 120/974 12
• CNF w/o equality 733/1290 57
• In total 204 successes of nonzero rating.
• Highest rated 4 problems of rating 0.78

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Conclusions

• Sound and complete fully automated first order
  theorem proving method
• Some encouraging results
• Offers a way to derive formulae with universal
  variables

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Future work

• Equality reasoning
• Equivalence relation approximation
• Model generation
• More optimizations